Problem: Simplify and expand the following expression: $ \dfrac{5}{3q + 24}+ \dfrac{3}{q + 10}+ \dfrac{3q}{q^2 + 18q + 80} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{5}{3q + 24} = \dfrac{5}{3(q + 8)}$ We can factor the quadratic in the third term: $ \dfrac{3q}{q^2 + 18q + 80} = \dfrac{3q}{(q + 8)(q + 10)}$ Now we have: $ \dfrac{5}{3(q + 8)}+ \dfrac{3}{q + 10}+ \dfrac{3q}{(q + 8)(q + 10)} $ The least common multiple of the denominators is: $ 3(q + 8)(q + 10)$ In order to get the first term over $3(q + 8)(q + 10)$ , multiply by $\dfrac{q + 10}{q + 10}$ $ \dfrac{5}{3(q + 8)} \times \dfrac{q + 10}{q + 10} = \dfrac{5(q + 10)}{3(q + 8)(q + 10)} $ In order to get the second term over $3(q + 8)(q + 10)$ , multiply by $\dfrac{3(q + 8)}{3(q + 8)}$ $ \dfrac{3}{q + 10} \times \dfrac{3(q + 8)}{3(q + 8)} = \dfrac{9(q + 8)}{3(q + 8)(q + 10)} $ In order to get the third term over $3(q + 8)(q + 10)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{3q}{(q + 8)(q + 10)} \times \dfrac{3}{3} = \dfrac{9q}{3(q + 8)(q + 10)} $ Now we have: $ \dfrac{5(q + 10)}{3(q + 8)(q + 10)} + \dfrac{9(q + 8)}{3(q + 8)(q + 10)} + \dfrac{9q}{3(q + 8)(q + 10)} $ $ = \dfrac{ 5(q + 10) + 9(q + 8) + 9q} {3(q + 8)(q + 10)} $ Expand: $ = \dfrac{5q + 50 + 9q + 72 + 9q}{3q^2 + 54q + 240} $ $ = \dfrac{23q + 122}{3q^2 + 54q + 240}$